For $2 \times 2$ matrices $A = \begin{pmatrix} a_{11} & a_{12}\\ a_{21} & a_{22} \end{pmatrix}$ prove that: If row-vectors of $A$ are linearly dependent, then $\det(A)=0$

I'm not sure how to do this correct. I would start by calculating its determinant:

$\det(A)=a_{11}a_{22} - a_{21}a_{12}$

Now let $\det(A)=0$

Then $0=a_{11}a_{22} - a_{21}a_{12} \Leftrightarrow a_{11}a_{22}=a_{21}a_{12}$

So we have $a_{11}=\frac{a_{21}a_{12}}{a_{22}}; \,\,\,\ a_{12}=\frac{a_{11}a_{22}}{a_{21}};\,\,\,\, a_{21}=\frac{a_{11}a_{22}}{a_{12}};\,\,\,\ a_{22}=\frac{a_{21}a_{12}}{a_{11}}$

If we write this as a matrix, we have $$\begin{pmatrix} \frac{a_{21}a_{12}}{a_{22}} & \frac{a_{11}a_{22}}{a_{21}}\\ \frac{a_{11}a_{22}}{a_{12}} & \frac{a_{21}a_{12}}{a_{11}} \end{pmatrix}$$

But now I have a problem, I don't know how to continue :s

On paper I have created following linear system:

$$\text{I}: \lambda_1 \frac{a_{21}a_{12}}{a_{22}} + \lambda_2 \frac{a_{11}a_{22}}{a_{21}} = 0$$

$$\text{II}: \lambda_1 \frac{a_{11}a_{22}}{a_{12}} + \lambda_2 \frac{a_{21}a_{12}}{a_{11}}=0$$

I have tried to form for $\lambda_1$ and $\lambda_2$ hoping that both will be equal to zero so that I had linearly dependent row-vectors but no..

Is there maybe a better way of doing this or is my attempt completely wrong? How you do it good?

  • 2
    $\begingroup$ You seem to be trying to prove the converse of the title. Start by assuming that the rows are linearly dependent and then show that the determinant is zero. $\endgroup$ – carmichael561 May 12 '18 at 1:30

By definition of linear dependence the second row is a multiple of the first (or vice versa). This means that:

$$ {{a}_{11}}{{a}_{22}}-{{a}_{12}}{{a}_{21}}={{a}_{11}}\left( b{{a}_{12}} \right)-{{a}_{12}}\left( b{{a}_{11}} \right)=b\left( {{a}_{11}}{{a}_{12}}-{{a}_{12}}{{a}_{11}} \right)=0 $$

  • $\begingroup$ @Ali: You've proposed a sensible edit, but generally pointing out such minor changes to the OP for their judgement on how to Edit is preferred. $\endgroup$ – hardmath May 12 '18 at 1:43

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